Trigonometry — Foundation

Q1 (10)

(a) Similar-triangles proof (6) Right triangle $ABC$, right angle at $C$, hypotenuse $AB=c$, $BC=a$, $CA=b$. Drop altitude $CD\perp AB$, foot $D$.

• $\triangle ACD \sim \triangle ABC$ (both right-angled, share $\angle A$ → AA). (1) So $\dfrac{b}{c}=\dfrac{AD}{b}\Rightarrow b^2 = c\cdot AD$. (1½)
• $\triangle BCD \sim \triangle ABC$ (both right-angled, share $\angle B$ → AA). (1) So $\dfrac{a}{c}=\dfrac{BD}{a}\Rightarrow a^2 = c\cdot BD$. (1½)
• Add: $a^2+b^2 = c(AD+BD) = c\cdot c = c^2$. (1) ∎

(b) Converse (4) Statement: if in a triangle the square of the longest side equals the sum of squares of the other two, the angle opposite that side is $90°$. (2) Check $9,12,15$: $9^2+12^2 = 81+144 = 225 = 15^2$. (1) → right-angled (right angle opposite the side 15). (1)

Q2 (8)

(a) (3) with opposite $=$ side opposite $\theta$, adjacent, hypotenuse $h$: $\sin=\frac{opp}{h},\ \cos=\frac{adj}{h},\ \tan=\frac{opp}{adj}$; $\csc=\frac{h}{opp},\ \sec=\frac{h}{adj},\ \cot=\frac{adj}{opp}$. (½ each)

(b) (3) $\csc\theta=\frac{h}{opp}=1/\sin\theta$; $\sec\theta=1/\cos\theta$; $\cot\theta=1/\tan\theta$. (1½) $\dfrac{\sin\theta}{\cos\theta}=\dfrac{opp/h}{adj/h}=\dfrac{opp}{adj}=\tan\theta$; similarly $\cot\theta=\cos/\sin$. (1½)

(c) (2) $\sin\theta=\frac{7}{25}\Rightarrow$ opp $=7$, hyp $=25$, adj $=\sqrt{25^2-7^2}=\sqrt{576}=24$. (1) $\cos\theta=\frac{24}{25},\ \tan\theta=\frac{7}{24},\ \sec\theta=\frac{25}{24}$. (1)

Q3 (10)

(a) (6) Equilateral side 2, altitude splits base into 1 each; altitude $=\sqrt{2^2-1^2}=\sqrt3$. (2) For $30°$ (top half-angle): opp $=1$, hyp $=2$ → $\sin30°=\frac12$, $\cos30°=\frac{\sqrt3}{2}$. (2) For $60°$ (base angle): opp $=\sqrt3$, hyp $=2$ → $\sin60°=\frac{\sqrt3}{2}$, $\cos60°=\frac12$. (2)

(b) (3) Legs 1,1 → hyp $=\sqrt2$. $\sin45°=\cos45°=\frac{1}{\sqrt2}$, $\tan45°=\frac11=1$. (3)

(c) (1) $\tan90°=\frac{opp}{adj}$; as the angle → $90°$ the adjacent side → $0$, so the ratio grows without bound → undefined (division by zero). (1)

Q4 (8)

(a) (4) In right triangle the two acute angles are $\theta$ and $90°-\theta$. The side opposite $\theta$ is adjacent to $(90°-\theta)$ and vice-versa. (1) $\sin(90°-\theta)=\frac{opp\ to\ (90-\theta)}{h}=\frac{adj\ to\ \theta}{h}=\cos\theta$. (1½) $\tan(90°-\theta)=\frac{opp\ to\ (90-\theta)}{adj\ to\ (90-\theta)}=\frac{adj\ to\ \theta}{opp\ to\ \theta}=\cot\theta$. (1½)

(b) (4) $\cos55°=\sin35°$ ⇒ first term $=1$. (1½) $\cos60°\sec60°=1$. (1) $\tan70°=\cot20°=1/\tan20°$ ⇒ $\tan20°\tan70°=1$. (1) Total $=1+1-1=1$. (½)

Q5 (8)

(a) (1) Tower height $h$ at foot $F$; $Q$ between $P$ and $F$, $PQ=40$, elevations $30°$ at $P$, $60°$ at $Q$.

(b) (5) Let $QF=x$. $\tan60°=\frac{h}{x}\Rightarrow h=x\sqrt3$. (1) $\tan30°=\frac{h}{x+40}\Rightarrow h=\frac{x+40}{\sqrt3}$. (1) Equate: $x\sqrt3=\frac{x+40}{\sqrt3}\Rightarrow 3x=x+40\Rightarrow x=20$. (2) $h=20\sqrt3\approx34.64\text{ m}$. (1)

(c) (2) $QF=x=20\text{ m}$. (2)

Q6 (6)

(a) (2) $\sin\theta=\frac{opp}{hyp}$; the hypotenuse is the longest side of a right triangle, so opp $<$ hyp, giving $\sin\theta<1$. A value $1.2>1$ is impossible. (2)

(b) (4) In right triangle: $opp^2+adj^2=hyp^2$ (Pythagoras). (1) Divide by $hyp^2$: (1) $\left(\frac{opp}{hyp}\right)^2+\left(\frac{adj}{hyp}\right)^2=1$ (1) i.e. $\sin^2\theta+\cos^2\theta=1$. (1)

[
{"claim":"Q1b: 9,12,15 satisfies 9^2+12^2=15^2","code":"result = (9**2+12**2==15**2)"},
{"claim":"Q2c: sin=7/25 gives cos=24/25, tan=7/24, sec=25/24","code":"adj=sqrt(25**2-7**2); result = (adj==24) and (Rational(24,25)==Rational(24,25)) and (Rational(7,24)==Rational(7,24)) and (Rational(25,24)==Rational(25,24))"},
{"claim":"Q4b: expression equals 1","code":"expr = sin(rad(35))/cos(rad(55)) + cos(rad(60))*sec(rad(60)) - tan(rad(20))*tan(rad(70)); result = simplify(expr-1)==0"},
{"claim":"Q5: tower height 20*sqrt(3), QF=20","code":"x=symbols('x',positive=True); sol=solve(Eq(x*sqrt(3),(x+40)/sqrt(3)),x)[0]; h=sol*sqrt(3); result = (sol==20) and (simplify(h-20*sqrt(3))==0)"}
]